If a graph is plotted between ln k and `1//T` for the first order reaction, the slope of the straight line so obtained is given by

Options

A. `-(E_(a))/( R )`
B. `-(E_(a))/(2.303R)`
C. `-(2.303)/(E_(a)R)`
D. `-(E_(a))/(2.303)`

Explanation

To find the slope of the graph plotted between ln k and \( \frac{1}{T} \) for a first-order reaction, we can follow these steps:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation relates the rate constant \( k \) of a reaction to the temperature \( T \):
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin.

### Step 2: Take the Natural Logarithm
To linearize the Arrhenius equation, take the natural logarithm of both sides:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

### Step 3: Rearrange the Equation
Rearranging the equation gives us:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]
This equation is in the form of \( y = mx + c \), where:
- \( y = \ln k \),
- \( m = -\frac{E_a}{R} \) (the slope),
- \( x = \frac{1}{T} \),
- \( c = \ln A \) (the y-intercept).

### Step 4: Identify the Slope
From the rearranged equation, we can see that the slope of the line when plotting \( \ln k \) against \( \frac{1}{T} \) is:
\[ \text{slope} = -\frac{E_a}{R} \]

### Conclusion
Thus, the slope of the straight line obtained from the graph of \( \ln k \) versus \( \frac{1}{T} \) for a first-order reaction is given by:
\[ -\frac{E_a}{R} \]

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